Question

Determine whether the regression equation is statisicallt significant (a=.05) and complete the source table below.

9) Determine whether the regression equation is statistically significant (a = .05) and complete the source table below (10 pts). Source SS - df Ms Foot Significant? (Y/N) Regression Residual Total 10) Compute the standard error of prediction for the regression equation (4 pts) Se =

Answer 1

5)

X	Y	$X^2$	$Y^2$	XY
1	6	1	36	6
2	8	4	64	16
3	7	9	49	21
4	5	16	25	20
5	3	25	9	15
6	4	36	16	24
7	1	49	1	7
8	2	64	4	16
Total=36	36	204	204	125

X ΣΧ 3ς - = = = 4.5
$\bar{Y}=\frac{\sum Y}{n}=\frac{36}{8}=4.5$

$S_{XX}=\sum (X-\bar X)^2=\sum X^2 - n\bar X^2=204-\left ( 8\times 4.5^2 \right )=204-162=42$

$S_{YY}=\sum (Y-\bar Y)^2=\sum Y^2 - n\bar Y^2=204-\left ( 8\times 4.5^2 \right )=204-162=42$

$S_{XY}=\sum (X-\bar X)(Y-\bar Y)=\sum XY- n\bar X\bar Y=125-\left ( 8\times 4.5^2 \right )=125-162=-37$

$r_s=\frac{S_{XY}}{\sqrt{S_{XX}\times S_{YY}}}=\frac{-37}{\sqrt{42\times 42}}=\frac{-37}{42}=-0.88095$

5) Or you can calculate it quickly using this formula

X	Y	d= X-Y	$d^2$
1	6	-5	25
2	8	-6	36
3	7	-4	16
4	5	-1	1
5	3	2	4
6	4	2	4
7	1	6	36
8	2	6	36
Total=36	36		158

${\displaystyle r_{s}=1-{\frac {6\sum d_{i}^{2}}{n(n^{2}-1)}},}$

$r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{6\times 158}{8\times 63}=-0.88095$

6) $H_0: \rho =0 \ \ vs \ \ H_1:\rho \neq 0$

Under Ho:

Test statistic is:

$\frac{r_s\sqrt{n-2}}{\sqrt{1-r_s^2}}=\frac{-0.88095\times \sqrt{8-2}}{\sqrt{1-(-0.88095)^2}}=-4.56009$

$r_{crit}=t_{n-2,\frac{\alpha }{2}}=t_{6,\frac{0.05}{2}}=t_{6,0.025}=2.447$

Since t(cal) =|-4.56009| = 4.56009 > 2.447 i.e t(tab) we have sufficient evidence to reject Ho at 5% level of significance

Is rs Significant? Yes

The p-value is .003851.
The result is significant at p < .05.

9) $SS_{TOT}=S_{YY}=42$

$SS_{REG}=\frac{S_{XY}^2}{S_XX}=\frac{(-37)^2}{42}=32.59524$

$SS_{RES}=S_{YY}-\frac{S_{XY}^2}{S_XX}=42-\frac{(-37)^2}{42}=9.404762$

MS= MEAN SUM OF SQUARES = SS/df

	df	SS	MS	F	Significance F
Regression	1	32.59524	32.59524/1=32.59524	32.59524/9.404762=20.79494	0.00385032
Residual	6	9.404762	9.404762/6=1.56746		YES
Total	7	42

10) Standard error of regression:

standard deviation of Y = $s_Y=\sqrt{\frac{\sum Y^2-n\left ( \bar{Y} \right )^2}{n-1}}=\sqrt{\frac{204-(8\times 4.5^2)}{8-1}}=\sqrt{\frac{204-162}{7}}=\sqrt{6} \\ \\ s_Y=2.44949$

$R^2 = r^2=\left ( -0.88095 \right )^2=0.776077$

$Adjusted \ R^2 = 1-\frac{n-1}{n-2}\times (1-R^2)= 1-\frac{8-1}{8-2}\times (1-0.776077)= 1-\frac{7}{6}\times (0.223923)=1-0.261243=0.7387566$

Standard error of regression:

${\color{DarkBlue} \mathbf{s_e=\left ( \sqrt{1-adjusted \ R^2} \right )\times s_Y=\left (\sqrt{0.2612433} \right )\times 2.44949=1.25198}}$

$\sigma ^2=\frac{1}{n-2}\left ( S_{YY}-\frac{S_{XY}^2}{S_{XX}} \right )=\frac{1}{6}\left ( 42-\frac{(-37)^2}{42} \right )=1.56746$

$s_e(\beta )=\sqrt{\frac{\sigma ^2}{S_{XX}}}=\sqrt{\frac{1.56746}{42}}=0.193185$